Thermodynamics: PV Diagrams & the First Law (AP Physics 2)

On the AP Physics 2 equation sheet, W = -PΔV is the work done ON the gas and the first law is ΔU = Q + W. On a PV diagram, the area under the curve gives the size of the work: compression makes W positive, expansion makes W negative, and a constant-volume step does zero work.

AP Physics: Unit 9 (topics 9.4 The First Law of Thermodynamics). This guide covers Topic 9.4 in AP Physics 2 Unit 9 (Thermodynamics), a unit weighted at 15 to 18 percent of the multiple-choice section. Every sign convention here matches the AP Physics 2 equation sheet: W = -PΔV is work done on the gas and ΔU = Q + W.

What a PV diagram shows

A PV diagram plots a gas's pressure on the vertical axis against its volume on the horizontal axis, and every point on it is a complete state of the gas. Pick a point and you know PP and VV, and the ideal gas law PV=nRTPV = nRT then fixes the temperature too. A process (heating the gas, compressing it, letting it expand against a piston) appears as a path from one point to another.

The path matters. Two processes can connect the same start and end states while transferring different amounts of heat and work along the way. That is the core skill of AP Physics 2 Unit 9: read the path, then turn it into numbers for work, heat, and internal energy using the first law of thermodynamics.

The first law of thermodynamics, AP style

The first law is energy conservation applied to a gas:

ΔU=Q+W\Delta U = Q + W

Here ΔU\Delta U is the change in the gas's internal energy, QQ is the heat added to the gas (positive in, negative out), and WW is the work done on the gas.

That last definition is the trap. The AP Physics 2 equation sheet prints W=−PΔVW = -P\Delta V, which is the work done ON the gas, and pairs it with ΔU=Q+W\Delta U = Q + W. Many textbooks instead define work as the work done BY the gas and subtract it in the first law. Same physics, opposite bookkeeping. On the exam, commit to the sheet's convention and the signs stay consistent.

Quick logic check: compress a gas (push the piston in) and you add energy to it, so WW is positive. Let it expand and it spends energy pushing the piston outward, so WW is negative.

Work equals area under the curve

On a PV diagram, the area between the process path and the volume axis equals the magnitude of the work, ∣W∣|W|. The sign comes from which way the path runs:

  • Path moves right (expansion): the gas pushes its surroundings outward, so the work done by the gas is positive and the work done on the gas is negative.
  • Path moves left (compression): the surroundings push the gas inward, so WW, the work done on the gas, is positive.
  • Path runs straight up or down (constant volume): zero area, so W=0W = 0.

The equation W=−PΔVW = -P\Delta V applies only when pressure is constant, because then the region under the path is a rectangle of height PP and width ∣ΔV∣|\Delta V|. When pressure changes, get the work from the area itself: count grid squares or split the shape into rectangles and triangles. And if a question asks for the work done by the gas, compute WW and flip the sign.

Isobaric, isochoric, and isothermal processes

Three process types cover most AP problems:

ProcessWhat is constantShape on the PV diagramWork and first law
IsobaricPressureHorizontal lineW=−PΔVW = -P\Delta V; heat and work both flow
IsochoricVolumeVertical lineW=0W = 0, so ΔU=Q\Delta U = Q
IsothermalTemperatureCurve with PVPV constantΔU=0\Delta U = 0 (ideal gas), so Q=−WQ = -W

Isochoric is also called isovolumetric. The isothermal result follows from kinetic theory: an ideal gas's internal energy depends only on its temperature, so constant temperature means constant internal energy. Compress a gas isothermally and the work you do on it (W>0W > 0) leaves as heat (Q<0Q < 0). One more label worth knowing: an adiabatic process exchanges no heat, so Q=0Q = 0 and ΔU=W\Delta U = W, which means the temperature changes even though no heat flows.

Cycles and net work

A thermodynamic cycle is a closed loop on the PV diagram: the gas ends in exactly the state where it began. Internal energy is a state function, meaning it depends only on the current state and not on the path taken, so over one full cycle ΔU=0\Delta U = 0 and the first law reduces to Q=−WQ = -W.

The magnitude of the net work for a cycle equals the area enclosed by the loop. The direction of travel sets the sign. A clockwise loop expands at high pressure and gets compressed at low pressure, so the gas does more work than it receives: net work done on the gas is negative and the gas absorbs net heat. That is a heat engine. A counterclockwise loop reverses everything: net positive work is done on the gas and it expels net heat, which is how a refrigerator behaves.

A sign-convention checklist

Most lost points in Unit 9 are sign errors, not physics errors. Run this checklist on every problem:

  • Decide up front whether the question wants work done on the gas or by the gas, and write down which one you are computing.
  • Find ΔV=Vf−Vi\Delta V = V_f - V_i before touching the work equation. Compression gives negative ΔV\Delta V, which makes W=−PΔVW = -P\Delta V positive.
  • Use W=−PΔVW = -P\Delta V only on constant-pressure segments. Everything else needs area.
  • Translate wording into signs: "the gas absorbs 400 J of heat" means Q=+400Q = +400 J, while "the gas loses 400 J" means Q=−400Q = -400 J.
  • On a full cycle, write ΔU=0\Delta U = 0 first. It usually cracks the problem open.

The same energy bookkeeping runs through conservation of energy and the work-energy theorem, and pressure itself is built up in the fluids guide.

Isobaric compression: finding W and the change in internal energy

A piston compresses a gas at a constant pressure of 2.0×105 Pa2.0 \times 10^5\ \text{Pa} from a volume of 6.0×10−3 m36.0 \times 10^{-3}\ \text{m}^3 to 2.0×10−3 m32.0 \times 10^{-3}\ \text{m}^3. During the compression the gas loses 500 J of heat. Find the work done on the gas and the change in its internal energy.

  1. Find the volume change: ΔV=Vf−Vi=2.0×10−3 m3−6.0×10−3 m3=−4.0×10−3 m3\Delta V = V_f - V_i = 2.0 \times 10^{-3}\ \text{m}^3 - 6.0 \times 10^{-3}\ \text{m}^3 = -4.0 \times 10^{-3}\ \text{m}^3. Negative, as expected for a compression.

  2. Apply the AP work equation: W=−PΔV=−(2.0×105 Pa)(−4.0×10−3 m3)=+800 JW = -P\Delta V = -(2.0 \times 10^5\ \text{Pa})(-4.0 \times 10^{-3}\ \text{m}^3) = +800\ \text{J}. The piston does positive work on the gas.

  3. Assign the heat a sign. The gas loses 500 J, so Q=−500 JQ = -500\ \text{J}.

  4. Apply the first law: ΔU=Q+W=−500 J+800 J=+300 J\Delta U = Q + W = -500\ \text{J} + 800\ \text{J} = +300\ \text{J}.

The work done on the gas is +800 J+800\ \text{J} and ΔU=+300 J\Delta U = +300\ \text{J}. The internal energy (and therefore the temperature) rises even though the gas is losing heat, because the compression adds energy faster than the heat leaves.

A two-step process: isobaric expansion, then isochoric cooling

A gas expands at a constant pressure of 1.5×105 Pa1.5 \times 10^5\ \text{Pa} from 2.0×10−3 m32.0 \times 10^{-3}\ \text{m}^3 to 5.0×10−3 m35.0 \times 10^{-3}\ \text{m}^3 (step 1), then cools at constant volume until its pressure drops to 0.50×105 Pa0.50 \times 10^5\ \text{Pa} (step 2). The gas absorbs a net 650 J of heat across both steps. Find the total work done on the gas and the change in its internal energy.

  1. Step 1 is isobaric: W1=−PΔV=−(1.5×105 Pa)(5.0×10−3 m3−2.0×10−3 m3)=−(1.5×105 Pa)(3.0×10−3 m3)=−450 JW_1 = -P\Delta V = -(1.5 \times 10^5\ \text{Pa})(5.0 \times 10^{-3}\ \text{m}^3 - 2.0 \times 10^{-3}\ \text{m}^3) = -(1.5 \times 10^5\ \text{Pa})(3.0 \times 10^{-3}\ \text{m}^3) = -450\ \text{J}. The expanding gas does work on its surroundings, so the work done on it is negative.

  2. Step 2 is isochoric: ΔV=0\Delta V = 0, so W2=0W_2 = 0. On the PV diagram this segment is a vertical line with no area under it.

  3. Total the work on the gas: W=W1+W2=−450 J+0=−450 JW = W_1 + W_2 = -450\ \text{J} + 0 = -450\ \text{J}.

  4. Apply the first law with Q=+650 JQ = +650\ \text{J}: ΔU=Q+W=+650 J+(−450 J)=+200 J\Delta U = Q + W = +650\ \text{J} + (-450\ \text{J}) = +200\ \text{J}.

The total work done on the gas is −450 J-450\ \text{J} (equivalently, the gas does 450 J of work on its surroundings) and its internal energy increases by 200 J200\ \text{J}.

Net work around a rectangular cycle

A gas runs clockwise around a rectangular cycle: it expands from 2.0×10−3 m32.0 \times 10^{-3}\ \text{m}^3 to 6.0×10−3 m36.0 \times 10^{-3}\ \text{m}^3 at 3.0×105 Pa3.0 \times 10^5\ \text{Pa}, drops to 1.0×105 Pa1.0 \times 10^5\ \text{Pa} at constant volume, is compressed back to 2.0×10−3 m32.0 \times 10^{-3}\ \text{m}^3, then returns to the starting pressure at constant volume. Find the net work done on the gas and the net heat for one full cycle.

  1. Top leg (expansion at high pressure): Wtop=−(3.0×105 Pa)(+4.0×10−3 m3)=−1200 JW_{\text{top}} = -(3.0 \times 10^5\ \text{Pa})(+4.0 \times 10^{-3}\ \text{m}^3) = -1200\ \text{J}.

  2. Bottom leg (compression at low pressure): Wbottom=−(1.0×105 Pa)(−4.0×10−3 m3)=+400 JW_{\text{bottom}} = -(1.0 \times 10^5\ \text{Pa})(-4.0 \times 10^{-3}\ \text{m}^3) = +400\ \text{J}.

  3. Both vertical legs have ΔV=0\Delta V = 0, so each contributes W=0W = 0.

  4. Add the legs: Wnet=−1200 J+400 J=−800 JW_{\text{net}} = -1200\ \text{J} + 400\ \text{J} = -800\ \text{J}. Check with the enclosed area: (3.0×105 Pa−1.0×105 Pa)(4.0×10−3 m3)=800 J(3.0 \times 10^5\ \text{Pa} - 1.0 \times 10^5\ \text{Pa})(4.0 \times 10^{-3}\ \text{m}^3) = 800\ \text{J}, which matches the magnitude.

  5. Over a full cycle ΔU=0\Delta U = 0, so Q=ΔU−W=0−(−800 J)=+800 JQ = \Delta U - W = 0 - (-800\ \text{J}) = +800\ \text{J}.

The net work done on the gas is −800 J-800\ \text{J} and the net heat absorbed is +800 J+800\ \text{J}. The gas turns net heat input into work output, so this clockwise cycle behaves as a heat engine.

Frequently asked questions

Why does the AP equation sheet write W = -PΔV with a minus sign?

Because on the AP Physics 2 sheet, W means the work done on the gas. When a gas expands, ΔV is positive and the gas pushes outward on its surroundings, so the surroundings do negative work on it: W = -PΔV correctly comes out negative. When the gas is compressed, ΔV is negative and W comes out positive. Pair it with ΔU = Q + W and the signs take care of themselves.

Is the area under a PV curve the work done on the gas or by the gas?

The area gives only the magnitude of the work; the two versions have the same size and opposite signs. Read the direction of the path to assign the sign: during expansion the work done on the gas is negative (and the work done by the gas is positive), and during compression it is the reverse.

My textbook writes the first law with a minus sign. Which version should I use on the AP exam?

Use the equation sheet's version: ΔU = Q + W, where W = -PΔV is the work done on the gas. Textbooks that define W as the work done by the gas subtract it instead. The physics is identical, but mixing the two conventions inside one problem guarantees a sign error, so commit to the sheet's form and stay with it.

What is the difference between isobaric, isochoric, and isothermal processes?

Isobaric means constant pressure: a horizontal line on the PV diagram, with W = -PΔV. Isochoric (or isovolumetric) means constant volume: a vertical line, W = 0, so ΔU = Q. Isothermal means constant temperature: a curve along which PV stays constant, and for an ideal gas ΔU = 0, so heat in is balanced by work done by the gas.

Why is the change in internal energy zero for a complete cycle?

Internal energy is a state function: it depends only on the gas's current state, not on how the gas got there. A cycle returns the gas to its starting state, so U returns to its starting value and ΔU = 0. Heat and work are not state functions, which is why net Q and net W around the cycle can each be nonzero; the first law just forces them to be equal and opposite.